Integrating Sine-Odd, Cosine-Even Products

Introduction

When an integrand contains powers of both sine and cosine, the parity of the exponents often points to a useful method. In this tutorial, the sine exponent is odd and the cosine exponent is even. The odd sine power provides a factor that pairs naturally with the derivative of cosine.

The method is direct: save one factor of sine, convert every remaining sine factor into cosine using \(\sin^2(x)=1-\cos^2(x)\), and substitute \(u=\cos(x)\). Because \(du=-\sin(x)\,dx\), keeping track of a minus sign is essential.

When the Method Applies

Consider an indefinite integral of the form \(\int \sin^m(x)\cos^n(x)\,dx\), where the exponents are integers, \(m\) is odd and positive, and \(n\) is even and nonnegative. Write \(m=2k+1\) and \(n=2\ell\), where \(k\) and \(\ell\) are nonnegative integers. Then

The odd power \(2k+1\) allows us to separate one factor of \(\sin(x)\). The remaining sine power is even, so it can be rewritten entirely in terms of cosine. The substitution \(u=\cos(x)\) then turns the integral into a polynomial integral.

The Substitution and Its Sign

Start by separating one sine factor. The remaining sine exponent, \(2k\), is even, so apply the identity \(\sin^2(x)=1-\cos^2(x)\):

Now let \(u=\cos(x)\). Differentiating gives \(du=-\sin(x)\,dx\), or equivalently \(\sin(x)\,dx=-du\). This converts the integral into

The minus sign comes from the derivative of cosine. The expression \((1-u^2)^k u^{2\ell}\) is a polynomial in \(u\), so it can be expanded and integrated term by term.

General Formula and Justification

The binomial theorem gives \((1-u^2)^k=\sum_{j=0}^{k}(-1)^j\binom{k}{j}u^{2j}\). Multiplying by \(u^{2\ell}\), integrating each power, and restoring \(u=\cos(x)\) yields a general antiderivative.

Here \(C\) is an arbitrary constant of integration. Each denominator is positive, since \(\ell\) and \(j\) are nonnegative integers.

To justify the result, define the polynomial antiderivative \(P(u)=-\int (1-u^2)^k u^{2\ell}\,du\), where the integral on the right is taken as an ordinary polynomial antiderivative. By construction, \(P'(u)=-(1-u^2)^k u^{2\ell}\). The chain rule then gives

Using \(1-\cos^2(x)=\sin^2(x)\), this becomes \(\sin^{2k+1}(x)\cos^{2\ell}(x)\), the original integrand. Thus differentiating the proposed antiderivative recovers the integrand. This argument works for every real \(x\); it does not require cosine to have an inverse on the interval under consideration.

Worked Examples

A Reliable Step-by-Step Procedure

For an integral in this parity pattern, use the steps below. The key is to decide which factor to save before expanding anything.

A Common Sign Mistake

The most frequent error is replacing \(\sin(x)\,dx\) by \(du\) instead of \(-du\). For example, consider \(\int\sin^3(x)\cos^2(x)\,dx\). After saving one sine factor and setting \(u=\cos(x)\), the correct substitution is

An incorrect calculation drops the minus sign and writes

That gives \(\frac{\cos^3(x)}{3}-\frac{\cos^5(x)}{5}+C\), whose derivative is \(-\sin^3(x)\cos^2(x)\), not the required integrand. Keeping the minus sign instead gives

Differentiating this corrected answer gives \(\sin(x)\cos^2(x)-\sin(x)\cos^4(x)=\sin(x)\cos^2(x)(1-\cos^2(x))=\sin^3(x)\cos^2(x)\), as required. A derivative check quickly detects the lost-sign error.

Why the Parity Pattern Is Useful

Parity tells you which trigonometric factor can supply a convenient differential. An odd sine power leaves one \(\sin(x)\,dx\) after the other sine factors have been rewritten in terms of cosine. Since the derivative of \(\cos(x)\) is \(-\sin(x)\), this leftover factor matches the substitution exactly.

The even cosine power does not need to be converted. It becomes an ordinary even power of \(u\), and the rewritten sine factors become powers of \(1-u^2\). Consequently, the substituted expression is a polynomial, whose antiderivative follows from the power rule.

The method depends on the stated parity pattern. If the sine exponent is not odd, there may be no single sine factor left with an even remainder to rewrite in this way. If the cosine exponent is odd instead, a different choice of saved factor and substitution is appropriate. The important habit is to use the exponent pattern to guide the substitution rather than attempting to manipulate powers at random.